Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
vector<int> merge(vector<int> a, vector<int> b) {
  a.insert(a.begin(), b.begin(), b.end());
  sort(a.begin(), a.end());
  a.erase(unique(a.begin(), a.end()), a.end());
  while (a.size() > 10) a.pop_back();
  return a;
}
int n, m, q;
vector<int> has[100013];
int h[100013];
int up[20][100013];
vector<int> low[20][100013];
vector<int> adj[100013];
int lca(int a, int b) {
  if (h[a] > h[b]) swap(a, b);
  for (int i = 19; i >= 0; i--)
    if ((h[b] - h[a]) & (1 << i)) b = up[i][b];
  if (a == b) return a;
  for (int i = 19; i >= 0; i--)
    if (up[i][a] != up[i][b]) a = up[i][a], b = up[i][b];
  return up[0][a];
}
vector<int> go(int a, int d) {
  vector<int> res = has[a];
  for (int i = 19; i >= 0; i--)
    if (d & (1 << i)) res = merge(res, low[i][a]), a = up[i][a];
  return res;
}
void dfs(int x, int p = 0, int d = 0) {
  h[x] = d;
  up[0][x] = p;
  low[0][x] = has[p];
  sort(has[x].begin(), has[x].end());
  while (has[x].size() > 10) has[x].pop_back();
  for (int i : adj[x])
    if (i != p) dfs(i, x, d + 1);
}
void query(int u, int v, int a) {
  int l = lca(u, v);
  vector<int> res = merge(go(u, h[u] - h[l]), go(v, h[v] - h[l]));
  while (res.size() > a) res.pop_back();
  printf("%d", res.size());
  for (int i : res) printf(" %d", i);
  printf("\n");
}
int main() {
  scanf("%d%d%d", &n, &m, &q);
  for (int i = 1; i < n; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  for (int i = 1; i <= m; i++) {
    int c;
    scanf("%d", &c);
    has[c].push_back(i);
  }
  dfs(1);
  for (int i = 1; i < 20; i++)
    for (int j = 1; j <= n; j++) {
      up[i][j] = up[i - 1][up[i - 1][j]];
      low[i][j] = merge(low[i - 1][j], low[i - 1][up[i - 1][j]]);
    }
  for (int i = 0; i < q; i++) {
    int u, v, a;
    scanf("%d%d%d", &u, &v, &a);
    query(u, v, a);
  }
  return 0;
}